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Question

The value of nC012 nC123+ nC234++(1)n nCn(n+1)(n+2) is

A
1n+2
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B
1n+1
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C
1n+3
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D
1(n+2)(n+1)
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Solution

The correct option is A 1n+2
Let
S= nC012 nC123+ nC234++(1)n nCn(n+1)(n+2)
Considering,
(1+x)n= nC0+ nC1x+ nC2x2++ nCnxn
Integrating w.rt. x,
(1+x)n+1n+1= nC0x+ nC1x22++ nCnxn+1n+1+k
Putting
x=0k=1n+1
(1+x)n+1n+1= nC0x+ nC1x22++ nCnxn+1n+1+1n+1
Again integrating w.r.t. x,
(1+x)n+2(n+1)(n+2)= nC0x212+ nC1x323++ nCnxn+2(n+1)(n+2)+xn+1+k1
Putting
x=0k1=1(n+1)(n+2)
(1+x)n+2(n+1)(n+2)= nC0x212+ nC1x323++ nCnxn+2(n+1)(n+2)+xn+1+1(n+1)(n+2)
Now, putting x=1,
0=S1(n+1)+1(n+1)(n+2)S=1n+2

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